If $$f(x)=\left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^4 x & \sin ^2 2 x \end{array}\right|,$$ then $\frac{1}{5} f^{\prime}(0)=$ is equal to :
Solution
<p>$$\begin{aligned}
& \left|\begin{array}{ccc}
2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\
3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\
2 \cos ^4 x & 3+2 \sin ^2 4 x & \sin ^2 2 x
\end{array}\right| \\
& \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\
& \left|\begin{array}{ccc}
2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\
3 & 0 & -3 \\
0 & 3 & -3
\end{array}\right| \\
& \mathrm{f}(\mathrm{x})=45 \\
& \mathrm{f}^{\prime}(\mathrm{x})=0 \\
&
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Differentiation · Topic: Derivatives of Standard Functions
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